E. not enough information given to decide Gaussian surface #1

Slides:



Advertisements
Similar presentations
The electric flux may not be uniform throughout a particular region of space. We can determine the total electric flux by examining a portion of the electric.
Advertisements

Rank the electric fluxes through each Gaussian surface shown in the figure from largest to smallest. Display any cases of equality in your ranking.
Chapter 22: The Electric Field II: Continuous Charge Distributions
Applications of Gauss’s Law
Lecture 6 Problems.
Week #3 Gauss’ Law September 7, What’s up Doc?? At this moment I do not have quiz grades unless I get them at the last minute. There was a short.
© 2012 Pearson Education, Inc. A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2)
C. less, but not zero. D. zero.
A Charged, Thin Sheet of Insulating Material
Slide 1 Electric Field Lines 10/29/08. Slide 2Fig 25-21, p.778 Field lines at a conductor.
General Physics 2, Lec 5, By/ T.A. Eleyan 1 Additional Questions (Gauss’s Law)
Chapter 23 Gauss’s Law.
Nadiah Alanazi Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions.
Today’s agenda: Announcements. Gauss’ Law Examples. You must be able to use Gauss’ Law to calculate the electric field of a high-symmetry charge distribution.
a b c Gauss’ Law … made easy To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1)
General Physics 2, Lec 5, By/ T.A. Eleyan 1 Additional Questions (Gauss’s Law)
III.A 3, Gauss’ Law.
Gauss’s Law The electric flux through a closed surface is proportional to the charge enclosed The electric flux through a closed surface is proportional.
© 2012 Pearson Education, Inc. { Chapter 22 Applying Gauss’s Law.
Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures.
Dr. Jie ZouPHY Chapter 24 Gauss’s Law (cont.)
Physics 2112 Unit 4: Gauss’ Law
Chapter 22 Gauss’ Law.
Copyright © 2009 Pearson Education, Inc. Chapter 22 Gauss’s Law.
ELECTRICITY PHY1013S GAUSS’S LAW Gregor Leigh
Introduction: what do we want to get out of chapter 24?
Q22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses.
Tue. Feb. 3 – Physics Lecture #26 Gauss’s Law II: Gauss’s Law, Symmetry, and Conductors 1. Electric Field Vectors and Electric Field Lines 2. Electric.
Gauss’s Law (III) Examples.
Flux and Gauss’s Law Spring Last Time: Definition – Sort of – Electric Field Lines DIPOLE FIELD LINK CHARGE.
Copyright © 2009 Pearson Education, Inc. Applications of Gauss’s Law.
Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures.
24.2 Gauss’s Law.
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
Gauss’s Law Basic Concepts Electric Flux Gauss’s Law
Cultural enlightenment time. Conductors in electrostatic equilibrium.
Electric flux To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x “area”
Physics 212 Lecture 4 Gauss’ Law.
Chapter 23 Gauss’s Law Spring 2008.
Gauss’s Law Gauss’s law uses symmetry to simplify electric field calculations. Gauss’s law also gives us insight into how electric charge distributes itself.
Gauss’s Law ENROLL NO Basic Concepts Electric Flux
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
Chapter 22 Gauss’s Law.
Electric flux To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x “area”
Gauss’s Law Electric Flux
Gauss’ Law and Applications
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Gauss’s Law Electric Flux Gauss’s Law Examples.
Flux and Gauss’s Law Spring 2009.
Last Lectures This lecture Gauss’s law Using Gauss’s law for:
TOPIC 3 Gauss’s Law.
Flux Capacitor (Schematic)
An insulating spherical shell has an inner radius b and outer radius c
C. less, but not zero. D. zero.
Gauss’s Law Electric Flux
Chapter 22 Gauss’s Law HW 4: Chapter 22: Pb.1, Pb.6, Pb.24,
Chapter 23 Gauss’s Law.
Last Lectures This lecture Gauss’s law Using Gauss’s law for:
Question for the day Can the magnitude of the electric charge be calculated from the strength of the electric field it creates?
problem1 - Charge in a Cube
Quiz 1 (lecture 4) Ea
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
Phys102 Lecture 3 Gauss’s Law
Chapter 23 Gauss’s Law.
Physics for Scientists and Engineers, with Modern Physics, 4th edition
Electric flux To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x “area”
Chapter 22 Gauss’s Law The Study guide is posted online under the homework section , Your exam is on March 6 Chapter 22 opener. Gauss’s law is an elegant.
Gauss’s Law.
Example 24-2: flux through a cube of a uniform electric field
Chapter 23 Gauss’s Law.
Presentation transcript:

E. not enough information given to decide Gaussian surface #1 Q22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is +q greater. B. the same. C. less, but not zero. D. zero. E. not enough information given to decide Gaussian surface #1 Gaussian surface #2 Answer: B

A22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is +q greater. B. the same. C. less, but not zero. D. zero. E. not enough information given to decide Gaussian surface #1 Gaussian surface #2

E. both surface C and surface D Q22.2 Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? surface A B. surface B C. surface C D. surface D E. both surface C and surface D Answer: E

A22.2 Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? surface A B. surface B C. surface C D. surface D E. both surface C and surface D

A. the electric field points radially outward. Q22.3 A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is –3Q, and it is insulated from its surroundings. In the region a < r < b, A. the electric field points radially outward. B. the electric field points radially inward. C. the electric field is zero. D. not enough information given to decide Answer: C

A22.3 A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is –3Q, and it is insulated from its surroundings. In the region a < r < b, A. the electric field points radially outward. B. the electric field points radially inward. C. the electric field is zero. D. not enough information given to decide

A. points generally toward the outer surface of the conductor. Q22.4 Conductor A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity Cavity A. points generally toward the outer surface of the conductor. B. points generally away from the outer surface of the conductor. C. is zero. D. not enough information given to decide Answer: B

A22.4 Conductor A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity Cavity A. points generally toward the outer surface of the conductor. B. points generally away from the outer surface of the conductor. C. is zero. D. not enough information given to decide

Just beneath the surface of this region, the electric field Q22.5 There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field A. points outward, toward the surface of the conductor. B. points inward, away from the surface of the conductor. C. points parallel to the surface. D. is zero. E. not enough information given to decide Answer: D

A22.5 There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field A. points outward, toward the surface of the conductor. B. points inward, away from the surface of the conductor. C. points parallel to the surface. D. is zero. E. not enough information given to decide

A. a uniformly charged sphere of radius R Q22.6 For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field? A. a uniformly charged sphere of radius R B. a spherical shell of radius R with charge uniformly distributed over its surface C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface E. Gauss’s law would be useful for finding the electric field in all of these cases. Answer: C

A22.6 For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field? A. a uniformly charged sphere of radius R B. a spherical shell of radius R with charge uniformly distributed over its surface C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface E. Gauss’s law would be useful for finding the electric field in all of these cases.